Supremum norm of a function whose Fourier transform is in weighted $L_2$

Question

Suppose that I have a positive function $w(x)>0$ and it defines a weighted $L^w_2$-space by $$ L^w_2 = \{f:{\mathbb R}\to {\mathbb C}\mid \int_{-\infty}^\infty w(x)|f(x)|^2dx<\infty\} $$ with norm $$ \|f\|_{L^w_2} = \sqrt{\int_{-\infty}^\infty w(x)|f(x)|^2dx} $$ Let $\Omega = [a,b]$ some finite interval. How can I bound $$ \sup_{\|\mathcal{F}[f]\|_{L^w_2}\leq 1, \|f\|_{L_2([a,b])}\leq 1}\|f\|_{L_\infty([a,b])} $$ in terms of $w(x)$? Here $\mathcal{F}[f]$ is Fourier transform of $f$.

Answer 1

The conclusion depends on the growth of the weight and does not hold for any weight.

For example let $w(x)\equiv 1$ and $$f_0(x)=\begin{cases}x^{-1/4}& 0<x<1/4\\ 0& {\rm otherwise} \end{cases}$$ Then $\|\mathcal{F}[f_0]\|_2=\|f_0\|_2=1,$ but the function $f_0$ is unbounded on the interval $[0,1/4].$

If the weight has sufficient growth at infinity the conclusion is valid. For example let $w(x)=1+x^2.$ Then the function $\mathcal{F}[f]$ is absolutely integrable as by the Cauchy-Schwarz inequality we get $$\displaylines{\int\limits_{-\infty}^\infty |\mathcal{F}[f](x)|\,dx=\int\limits_{-\infty}^\infty |\mathcal{F}[f](x)|w(x)^{1/2}\,w(x)^{-1/2}\,dx \\ \le \pi^{1/2}\,\|\mathcal{F}[f]\|_{L_2^w}}$$ By the Fourier inversion formula we get that $f$ is continuous and $$f(t)=\int\limits_{-\infty}^\infty \mathcal{F}[f](x)\,e^{2\pi itx}\,dx$$ Therefore $$\|f\|_\infty\le \int\limits_{-\infty}^\infty |\mathcal{F}[f](x)|\,dx\le \pi^{1/2}\,\|\mathcal{F}[f]\|_{L_2^w}$$ By the same reasoning if $w(x)^{-1}$ is integrable then $$\|f\|_\infty\le \|w^{-1}\|_1^{1/2}\|\,\|\mathcal{F}[f]\|_{L_2^w}$$